Let $V$ be a finite-dimensional vector space over a field $k$, say of characteristic $0$. The symmetric group $S_n$ acts on the tensor power $V^{\otimes n}$ in the obvious way, and this action defines two subspaces of $V^{\otimes n}$, the subspace on which $S_n$ acts via the trivial character and the subspace on which $S_n$ acts via the antisymmetric character.

Question 0: Is the construction of these subspaces functorial in $V$? If it is, are the corresponding functors naturally isomorphic to the symmetric and exterior powers, and if that's true, are the corresponding natural isomorphisms unique?

If the answers to Question 0 turn out more or less like I suspect, we should not regard these subspaces as completely synonymous with the symmetric power $S^n V$ and the exterior power $\Lambda^n V$, respectively, since these are naturally thought of as quotients of $V^{\otimes n}$. (This issue recently came up in another MO question.)

Question 1: Is there an established notation in the literature which respects this distinction?

$\left(\otimes V\right)_{\mathrm{symm}}$ and $\left(\otimes V\right)_{\mathrm{anti}}$ are what I have encountered. Of course, they don't look particularly slick.
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darij grinbergFeb 7 '11 at 21:27

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And of course they are functorial. Just check that the image of $\left(\otimes^n V\right)_{\mathrm{symm}}$ under the $n$-th tensor power of a linear map $f:V\to W$ is included in $\left(\otimes^n W\right)_{\mathrm{symm}}$. (That's because the action of $S_n$ is functorial in $V$.)
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darij grinbergFeb 7 '11 at 21:28

@Qiaochu: I believe that there are problems in characteristic p (although I could be wrong). I'm also pretty sure that the splitting is not unique, in any case.
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Harry GindiFeb 7 '11 at 21:29

I am pretty sure they are not isomorphic to the symmetric power / the exterior power if the characteristic of $k$ is bad. (Several counterexamples were posted here.)
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darij grinbergFeb 7 '11 at 21:30

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@darij You shouldn't be answering the question in the comments. It makes it difficult to clear the question from MathOverflow without looking like you're stealing credit from the actual answerer.
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Greg KuperbergFeb 7 '11 at 21:32

3 Answers
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This only answers part of your question 0 unfortunatly. The construction is certainly functorial, but the two notions of symmetric/alternating power do not always agree. Let's write $\operatorname{Sym}^n (V)$ for the symmetric tensors, and $\operatorname{Alt} ^n (V)$ for the alternating tensors. I wish this were established notation, but it probably isn't. Let $p$ be the characteristic of the field. Note $V^{\otimes n}$ is a $kGL(V) - \Sigma_n$ bimodule ($\Sigma_n$ is the symmetric group). Then if $r$ is less than $p$, or if $p=0$, $\operatorname{Sym}^r (V) \cong S^r(V)$ and $\operatorname{Alt}^r (V) \cong \Lambda ^r(V)$ as $GL(V)$-modules (this is proved by writing down maps explicitly).

If $r \geq p$ then $S^r$ and $\operatorname{Sym}^r$ are the contravariant (i.e. transpose) duals of one another as $GL$ modules. I imagine the same is true of the alternating power/antisymmetric tensors.

Let $V^{\otimes n}_{\mathrm{symm}}$ denote the subspace of the tensor power $V^{\otimes n}$ consisting of the elements on which $S_n$ acts trivially. Let $V^{\otimes n}_{\mathrm{alt}}$ denote the subspace of the tensor power $V^{\otimes n}$ consisting of the elements on which $S_n$ acts by the sign representation.`

Of course, multiplying such an isomorphism by a scalar $\neq 0$ yields another isomorphism. This is all the freedom we have: any two isomorphisms between the functor $V\mapsto V^{\otimes n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$ and the functor $V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$ are equal up to scalar, and similarly for the other pair of functors.

To prove this, we let $P$ be an isomorphism from the functor $V\mapsto V^{\otimes n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$ to the functor $V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$. Let $\lambda\in k$ be defined by $P_k\left(1\otimes 1\otimes ...\otimes 1\right)=\lambda 1\cdot 1\cdot ...\cdot 1$, where $P_k$ is the isomorphism $P$ at the object $V=k$, and $\cdot$ denotes the multiplication in the symmetric algebra (because it is commutative).

Since $k$ has characteristic $0$, the space $ V^{\otimes n}_{\mathrm{symm}}$ is generated by the tensors $v\otimes v\otimes\dots\otimes v$ for $v\in V$. (This is Lemma 7 from Crawley-Boevey's above-mentioned text, sent back to $V^{\otimes n}_{\mathrm{symm}}$ from $\mathrm{S}^n\left(V\right)$.) We are now going to prove that $P_V\left(v\otimes v\otimes \ldots\otimes v\right)=\lambda v\cdot v\cdot \ldots\cdot v$ for every $v\in V$.`

In order to show this, let $f:k\to V$ be a vector space homomorphism given by $f\left(1\right)=v$. The functoriality of $P$ now yields

This rewrites as $P_V\left(v\otimes v\otimes \ldots\otimes v\right)=\lambda v\cdot v\cdot \ldots\cdot v$, and we are done.

This yields (since the space $ V^{\otimes n}_{\mathrm{symm}}$ is generated by the tensors $v\otimes v\otimes\ldots\otimes v$ for $v\in V$) that the map $P_V$ is just the canonical projection from $ V^{\otimes n}_{\mathrm{symm}}$ to $\mathrm{S}^n\left(V\right)$, multiplied with the scalar $\lambda$. Since $\lambda$ does not depend on $V$, this shows us that our isomorphism $P$ from the functor $V\mapsto V^{\otimes n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$ to the functor $V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$ is the projection isomorphism times $\lambda$. In other words, all the freedom we have to choose this isomorphism is the freedom of choosing the scalar factor to multiply with. The same argument works for the other pair of functors.

Oh, and if you are wondering why the whole text is in a "code" section: I can't be bothered to put every single formula that contains indices in backticks, and there doesnt seem to be a good way to figure out whether this is necessary just by looking at the TeX code. We need an auto-backtick-tex bot...
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darij grinbergFeb 7 '11 at 22:37

As Qiaochu's MathOverflow source shows, there is no problem, or no problem any longer, with underscores. You only need the backticks at the most for asterisks, and Qiaochu's solution is to use \ast.
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Greg KuperbergFeb 7 '11 at 22:49

I, on the other hand, can't handle reading in the giant "code" font, so I have taken the liberty to add the back-ticks where needed. I didn't back-tick everything, so I hope I didn't miss a \{.
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Theo Johnson-FreydFeb 8 '11 at 3:25